Okies here's the solution:
The word CONSTANTINOPLE has 9 consonants of which N repeates 3 times, T repeats 2 times and 5 vowels of which O repeates 2 times.
1. 14 letters can be arranged 14! ways and there are some characters repeating so their arramgments are double counted.
So Ans = 14! / 3! * 2! *2!.
2. Vowels and consonants are to be considered to seperate groups. So both can be aranged in 2! ways.
Also consonants can be arranged in 9! / 3! * 2! ways and vowels can be arranged in 5!/ 2! ways.
So Ans = 2 * 9! * 5! / 3! * 2! * 2!
3. if no vowels are together than we start how many places are there for vowels to be there.
There are 9 consonants which can be arranged amongst themselves in 9!/ 3! * 2! ways.
So between the consonants there is places for vowels and there are 10 such places.
Consider this.
_C_C_C_C_C_C_C_C_C_
So, in 10 places vowels can be arranged in 10P5 ways that is 10!/5!. Also there are 2 vowels which are repeated so final ways is 10! / 5! * 2!
So Ans = 10! * 9! / 5! * 3! * 2! *2!
4. Probability of getting a word with NO two vowels is then
Ans 3 / Ans 1.
So Ans = 18/143.